[Quantum Mechanics][정리중][Modifying]Simple Harmonic Oscillator

Classical Harmonic Oscillator

Recall Hooke’s Law restoring force \(F\) is proportional to the displacement ‘\(x\)’. \[F=-kx\]

Where k is the force constant(spring constant) and the negative sign denote that restoring force occur to the opposite direction to the displacement ‘\(x\)’.

By applying Newton’s second law we can get \[F=ma=m\frac{d^2x}{dt^2}=-kx\] which it will lead us to have \[m\frac{d^2x}{dt^2}+kx=0\]

This is the case where external force is equal to zero. (\(F_{ext}=0\)) The general solution for differential equation is \[Acos(w_0t+\phi_0)+Bsin(w_0t+\phi_0)\]

Where ‘\(w_0=\sqrt{\frac{k}{m}}\)’ which is natural frequency of the oscillator. However, for the simplicity let us choose solution ‘\(x\)’ is equal to \[x=x_0sin(wt+\phi_0)\]

Furthermore, let us choose initial condition, so that ‘\(\phi_0=0\)’. Then the solution would be \[x(t)=x_0sin(wt)\]

When we take derivative of distance over time, we can get velocity \[v(t)=\frac{dx}{dt}=x_0wcos(wt)\]

Finally, we can get energy \(E\) in which \[E=\frac{mv^2(t)}{2}+\frac{1}{2}mw^2x^2(t)\]

Where ‘\(k=mw^2\)’ when we plug ‘\(x(t)\)’ and ‘\(v(t)\)’ in energy equation we can get \[E=\frac{m}{2}x_0^2w^2cos^2(wt)+\frac{m}{2}x_0^2w^2sin^2(wt)\]

Since ‘\(sin^2(x)+cos^2(x)=1\)’ we can simplify in to \[E=\frac{m}{2}x_0^2w^2\]

Quantum Harmonic Oscillator

Solution:
a. The first order shift vanishes.

\[\Delta_0^1 = \la 0 | bx | 0 \ra = 0\]

The second order shift is

\[\Delta_0^2 = -\f{b^2}{2m\omega^2}\]

This comes from

\[\Delta_0^2 = \sum_{k\neq n} \f{|V_{k0}|^2}{E_0^0 - E_k^0} = \f{b^2|\la 1|x|0\ra|^2}{E_0^0 - E_1^0}\]

Bring in a negative sign and switch around the denominator

\[\Delta_0^2 = -\f{b^2|\sqrt{\f{\hbar}{2m\omega}}(\sqrt{1}\delta_1^1)|^2}{\frac{3}{2}\hbar\omega - \frac{1}{2}\hbar\omega}\]

\[\Delta_0^2 = -\f{b^2\hbar}{2m\omega\hbar\omega} = -\f{b^2}{2m\omega^2}\]

b. The quickest way to do this is to add and subtract some constant from the Hamiltonian.

\[H = \f{p^2}{2m} + \f{1}{2}m\omega^2x^2 + bx + K - K\]

But this constant will be picked very conveniently so as to get rid of this ugly \(bx\) term. I want this form:

\[H = \f{p^2}{2m} + \f{1}{2}m\omega^2(x + C)^2 - K\]

What does this get me? What should \(C\) be equal to in terms of \(b\)?

\[H = \f{p^2}{2m} + \f{1}{2}m\omega^2(x^2 + 2Cx + C^2) - K\]

So \(C\) must be

\[\f{1}{2}m\omega^2 2C x = b x \rightarrow C = \f{b}{m\omega^2}\]

and we

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